The chemical equation would be:
2NO(g) + O2(g) --> 2NO2 (g)
<span>At equilibrium state, the partial pressure of the gases would be as follows : </span>
<span>NO = 522 - 2x </span>
<span>O2 = 421 - x </span>
<span>NO2 = 2x </span>
<span>- - - - - - - - - - - - -</span>
<span>943 - x = 748 </span>
<span>x = 195</span>
Calculating for Kp,
<span>Kp = (NO2)^2/ ((NO)^2 * (O2)) </span>
<span>Kp = (2 * 195)^2/ ((522 - 2 * 195)^2 * (421 - 195)) </span>
<span>Kp = 0.0386 </span>
Because the amount of hydrogens all depends on the valence electrons
In order to find the two statements, we must first define what the enthalpy of formation and the enthalpy of reaction mean.
Enthalpy of formation:
The change in enthalpy when one mole of substance is formed from its constituent elemetns at standard state.
Enthalpy of reaction:
The change in enthalpy when a reaction occurs and the reactants and products are in their standard states.
Now, we check the statements. The true ones are:
The Hrxn for C(s) + O₂(g) → CO₂(g) is the same as Hf for CO₂
This is true because the formation of carbon dioxide requires carbon and oxygen in their standard states.
The Hf for Br₂<span>(l) is 0 kJ/mol by definition.
Because the bromine is present in its standard state, the enthalpy of formation is 0.
</span><span>The Hrxn for the reaction 1.5H</span>₂<span>(g) + 0.5N</span>₂<span>(g) </span>→ <span>NH</span>₃<span>(g) is the same as the Hf for NH</span>₃<span>(g)
The reactants and products are present in their standard state, and the reaction is the same as the one occurring during the formation of ammonia.
</span>
Answer:
a. Kw = 1.0 × 10⁻¹⁴
Explanation:
a. Let's consider the self-ionization of water.
2 H₂O(l) ⇄ H₃O⁺(aq) + OH⁻
The ion-product of water (Kw) at 25 °C is:
Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴
c. Considering [H⁺] = [H₃O⁺] = 4.0 × 10⁻⁹ M, the concentration of OH⁻ is:
[OH⁻] = 1.0 × 10⁻¹⁴/[H₃O⁺] = 2.5 × 10⁻⁶ M
b. We can calculate the pOH using the following expression.
pOH = -log [OH⁻] = -log 2.5 × 10⁻⁶ = 5.6
d. We can calculate the pH using the following expression.
pH = -log [H⁺] = -log 4.0 × 10⁻⁹ = 8.4